Graph this system of equations and solve. $8x-2y = -10$ $y = -4 x - 3$ 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 Click and drag the points to move the lines.
Convert the first equation, $8x-2y = -10$ , to slope-intercept form. $y = 4 x + 5$ The y-intercept for the first equation is $5$ , so the first line must pass through the point $(0, 5)$ The slope for the first equation is $4$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move up $1$ position to the right. $4$ positions up from $(0, 5)$ is $(1, 9)$ Graph the blue line so it passes through $(0, 5)$ and $(1, 9)$ The y-intercept for the second equation is $-3$ , so the second line must pass through the point $(0, -3)$ The slope for the second equation is $-4$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move down (because it's negative) $1$ position to the right. $4$ positions down from $(0, -3)$ is $(1, -7)$ Graph the green line so it passes through $(0, -3)$ and $(1, -7)$ The solution is the point where the two lines intersect. The lines intersect at $(-1, 1)$.